The maximum potential energy of a block executing simple harmonic motion is 25 J. A is amplitude of oscillation. At A2, the kinetic energy of the block is:

The maximum potential energy of a block executing simple harmonic… — Solution | JEE Physics

MCQEasyJEE 2023Energy in SHM

The maximum potential energy of a block executing simple harmonic motion is $25 \, \text{J}$ . $A$ is amplitude of oscillation. At $\frac{A}{2}$ , the kinetic energy of the block is:

A
$37.5 \, \text{J}$
B
$9.75 \, \text{J}$
C
$18.75 \, \text{J}$
D
$12.5 \, \text{J}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The maximum potential energy in SHM is $25 \, \text{J}$ .

Find: The kinetic energy at displacement $x = \frac{A}{2}$ .

For simple harmonic motion,

U_{\max} = \frac{1}{2} m \omega^2 A^2 = 25 \, \text{J}

The kinetic energy at $x = \frac{A}{2}$ is written in the solution as

KE = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 \left(A^2 - \left(\frac{A}{2}\right)^2\right)

So,

KE = \frac{1}{2} m \omega^2 \left(A^2 - \frac{A^2}{4}\right)

KE = \frac{1}{2} m \omega^2 \cdot \frac{3A^2}{4}

KE = \frac{3}{4} \left(\frac{1}{2} m \omega^2 A^2\right)

Using $\frac{1}{2} m \omega^2 A^2 = 25 \, \text{J}$ ,

KE = \frac{3}{4} \times 25 = 18.75 \, \text{J}

Therefore, the kinetic energy is $18.75 \, \text{J}$ . The solution working gives option C, although the solution incorrectly says A.

Energy relation in SHM

Given: Maximum potential energy is $25 \, \text{J}$ .

Find: Kinetic energy at $\frac{A}{2}$ .

In SHM, total mechanical energy remains constant and equals the maximum potential energy:

E = U_{\max} = 25 \, \text{J}

Potential energy at displacement $x$ is

U = \frac{1}{2} m \omega^2 x^2

At $x = \frac{A}{2}$ ,

U = \frac{1}{2} m \omega^2 \left(\frac{A}{2}\right)^2 = \frac{1}{4}\left(\frac{1}{2} m \omega^2 A^2\right)

Since $\frac{1}{2} m \omega^2 A^2 = 25 \, \text{J}$ ,

U = \frac{25}{4} = 6.25 \, \text{J}

Hence,

KE = E - U = 25 - 6.25 = 18.75 \, \text{J}

Therefore, the correct option is C.

Common mistakes

Using the maximum potential energy as the kinetic energy at $x = \frac{A}{2}$ is incorrect because maximum potential energy occurs only at the extreme position. At $\frac{A}{2}$ , energy is shared between kinetic and potential parts. First find the potential energy at that displacement, then subtract from total energy.
Assuming potential energy varies linearly with displacement is wrong. In SHM, $U \propto x^2$ , not $x$ . So at $x = \frac{A}{2}$ , the potential energy becomes one-fourth of the maximum value, not one-half.
Reading the solution 'The Correct Option is A' without checking the working leads to the wrong answer. The extracted solution steps clearly evaluate the kinetic energy as $18.75 \, \text{J}$ , which matches option C. Always trust the actual derivation when a header conflicts with it.

Practice more Energy in SHM questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Energy in SHM

Answer

Step-by-step solution

Common mistakes

Practice more Energy in SHM questions

Related questions